1. ## infinite series

Find the sum of the infinite series
$\displaystyle 1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....$

2. Originally Posted by ihmth
Find the sum of the infinite series
$\displaystyle 1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....$
Re-arrange:

$\displaystyle = (1/5 + 1/5^4 + 1/5^7 + ....) + (2/5^2 + 2/5^5 + 2/5^8 + .....) + (3/5^3 + 3/5^6 + 3/5^9 + .....)$

$\displaystyle = 1/5 (1 + 1/5^3 + 1/5^6 + ....) + 2/5^2 (1 + 1/5^3 + 1/5^6 + ....) + 3/5^3 (1 + 1/5^3 + 1/5^6 + ......)$

$\displaystyle = (1/5 + 2/5^2 + 3/5^3) (1 + 1/5^3 + 1/5^6 + ....)$

$\displaystyle = 38/5^3 (1 + 1/5^3 + 1/5^6 + ....)$

where $\displaystyle (1 + 1/5^3 + 1/5^6 + ....)$ is an infinite geometric series a = 1 and $\displaystyle r = 1/5^3$. And you know how to get that sum, right?

3. Originally Posted by ihmth
Find the sum of the infinite series
$\displaystyle 1/5 + 2/5^2 + 3/5^3 + 1/5^4 + 2/5^5 + 3/5^6 + 1/5^7 + 2/5^8 + 3/5^9 + .....$
What Mr Fantastic said, but also if you group the terms in threes you will
find its a geometric series and we all know how to sum those.

RonL

4. Hello, ihmth!

Yet another approach . . .

Find the sum of the infinite series:

. . $\displaystyle S \;=\;\frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \frac{1}{5^4} + \frac{2}{5^5} + \frac{3}{5^6} + \frac{1}{5^7} + \frac{2}{5^8} + \frac{3}{5^9} + \hdots$

Multiply both sides by $\displaystyle 5^3$

. . $\displaystyle 5^3S \;=\;5^2 + 2\!\cdot\!5 + 3 + \underbrace{\frac{1}{5} + \frac{2}{5^2} + \frac{3}{5^3} + \frac{1}{5^4} + \frac{2}{5^5} + \frac{3}{5^6} + \hdots}_{\text{This is }S}$

And we have: .$\displaystyle 125S \;=\;38 + S\quad\Rightarrow\quad 124S \:=\:38$

Therefore: .$\displaystyle \boxed{S \;=\;\frac{19}{62}}$

5. Originally Posted by mr fantastic
Re-arrange:

$\displaystyle = (1/5 + 1/5^4 + 1/5^7 + ....) + (2/5^2 + 2/5^5 + 2/5^8 + .....) + (3/5^3 + 3/5^6 + 3/5^9 + .....)$

$\displaystyle = 1/5 (1 + 1/5^3 + 1/5^6 + ....) + 2/5^2 (1 + 1/5^3 + 1/5^6 + ....) + 3/5^3 (1 + 1/5^3 + 1/5^6 + ......)$

$\displaystyle = (1/5 + 2/5^2 + 3/5^3) (1 + 1/5^3 + 1/5^6 + ....)$

$\displaystyle = 38/5^3 (1 + 1/5^3 + 1/5^6 + ....)$

where $\displaystyle (1 + 1/5^3 + 1/5^6 + ....)$ is an infinite geometric series a = 1 and $\displaystyle r = 1/5^3$. And you know how to get that sum, right?
Tnx